Limit of a quotient without using L'Hopital rule

Question

How is the limit of :

$$\lim_{x \rightarrow 0}\frac{\sin(ax)}{\sin(bx)}$$ found without using L'Hopital's rule. I tried substituting $\tan(ax)\cos(ax)$ for $\sin(ax)$, but did not get the answer.

Answer 1

$$\frac{\sin(ax)}{\sin(bx)}=\frac{\sin(ax)}{ax}\frac{bx}{\sin(bx)}\frac{a}{b}$$

Since $\frac{\sin(ax)}{ax}\to1$ and $\frac{bx}{\sin(bx)}\to1$ our limit tends to $a/b$.